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3 added 53 characters in body

I don't know that there is anything as nice as the way that the two main operations in a ring fit together. However there are ways to introduce many linked operations which, while mainly done for the purpose of doing so, are perhaps not too overly frivolous. Here is one:

Given a finite set $S$ with $k$ elements, there are $k^{k^2}$ binary operations and each operation $\diamond$ can be represented by its Cayley Table the $k \times k$ square with $i,j$ entry $i \diamond j$ (assuming a known order). This single operation defines a quasi-group if the table is a Latin Square, each symbol appears once in each row and column. Equivalently (and without an agreed order), for every $a,b$ there are unique $x,y$ with $x \diamond a=b$ and $a \diamond y=b.$ This defines two operations $y=a \backslash b$ and $x=b / a$ with $$b=a \diamond (a \backslash b)=a \backslash (a \diamond b)=(b /a) \diamond a=(b \diamond a) / a.$$ This equational definition of a quasi-group via 3 operations is available for infinite quasi-groups as well. If we introduce the projections $a\pi_1b=a$ and $a\pi_2b=b$ then we can write the previous equations as $$a\pi_2b=(a\pi_1b)\diamond(a\backslash b)=\cdots$$

Two Latin Squares (given by quasigroups $(S,\cdot)$ and $(S,\diamond)$) are orthogonal if the map $(x,y) \to (x\cdot y,x\diamond y)$ is a bijection from $S \times S$ to itself. This It is not known how many pairwise orthogonal quasi-groups (aka Mutually Orthogonal Latin Squares MOLS) are possible on a set of size $k.$ Certainly no more than $k-1$ and this is possible when (and perhaps only when) $k$ is a prime power.

So I suppose that a set of $t$ MOLS on the same set could be described as $t$ operations $\diamond_i$ whose individual quasi-group nature is described asserted (as before) by using another $2t$ operations $a \backslash_i b$ and $a/_i b$ (along with $\pi_1$ and $\pi_2$) and whose pairwise orthogonality is defined by specifieded using a further $t(t-1)$ operations $\leftarrow_{ij}$ and $\rightarrow_{ij}$ so that $x=$ x=a \leftarrow_{ij}b$and$y=a \rightarrow_{ij} b$satisfy$(x\diamond_iy,x\diamond_jy)=(a,b).$Or, if we have no shame, $$((a \leftarrow_{ij}b)\diamond_i(a \rightarrow_{ij} b),(a \leftarrow_{ij}b)\diamond_j(a \rightarrow_{ij} b))=(a\pi_1b,a\pi_2b).$$ On one hand$\leftarrow_{ij}=\rightarrow_{ji}$so maybe a total count of$t^2+2t+2$is too greedy. On the other hand, maybe a few more operations such as$a\diamond^i b=b\diamond_ia$could be shoehorned in. But I will stop there. 2 added 1462 characters in body I don't know that there is anything as nice as the way that the two main operations in a ring fit together. However there are ways to introduce many linked operations which, while mainly done for the purpose of doing so, are perhaps not too overly frivolous. Here is one: Given a finite set$S$with$k$elements, there are$k^{k^2}$binary operations and each operation$\diamond$can be represented by its Cayley Table the$k \times k$square with$i,j$entry$i \diamond j$(assuming a known order). This single operation defines a quasigroup quasi-group if the table is a Latin Square, each symbol appears once in each row and column. Equivalently (and without an agreed order), for every$a,b$there are unique$x,y$with$x \diamond a=b$and$a \diamond y=b.$This defines two operations$y=a \backslash b$and$x=a x=b / b$a$ with $$b=a \diamond (a \backslash b)=a \backslash (a \diamond b)=(b /a) \diamond a=(b \diamond a) / a.$$ This equational definition of a quasi-group via 3 operations is available for infinite quasi-groups as well. If we introduce the projections $k$ is a prime power (a\pi_1b=a$and perhaps only$a\pi_2b=b$then ) it is possible we can write the previous equations as $$a\pi_2b=(a\pi_1b)\diamond(a\backslash b)=\cdots$$ Two Latin Squares (given by quasigroups$(S,\cdot)$and$(S,\diamond)$) are orthogonal if the map$(x,y) \to have (x\cdot y,x\diamond y)$is a bijection from$k-1$S \times S$ to itself. This It is not known how many pairwise orthogonal quasi-groups (aka Mutually Orthogonal Latin Squares in that any two give all MOLS) are possible on a set of size $k^2$ k.$Certainly no more than$k-1$and this is possible ordered pairs when combined position by position(and perhaps only when)$k$is a prime power. This can go up to So I suppose that a set of$k+1$if combined with projection onto t$ MOLS on the first and second coordinates. That should same set could be possible to describe described as $k-1$ or t$operations$k+1$related \diamond_i$ whose individual quasi-group nature is described (as before) by another $2t$ operations $a \backslash_i b$ and $a/_i b$ (along with $\pi_1$ and $\pi_2$) and whose pairwise orthogonality is defined by a further $t(t-1)$ operations $\leftarrow_{ij}$ and $\rightarrow_{ij}$ so that $x=$ and $y=a \rightarrow_{ij} b$ satisfy $(x\diamond_iy,x\diamond_jy)=(a,b).$ Or, if we have no shame, $$((a \leftarrow_{ij}b)\diamond_i(a \rightarrow_{ij} b),(a \leftarrow_{ij}b)\diamond_j(a \rightarrow_{ij} b))=(a\pi_1b,a\pi_2b).$$

On one hand $\leftarrow_{ij}=\rightarrow_{ji}$ so maybe a total count of $t^2+2t+2$ is too greedy. On the other hand, maybe a few more operations could be shoehorned in. But I will stop there.

1

Given a finite set with $k$ elements there are $k^{k^2}$ binary operations and each operation $\diamond$ can be represented by its Cayley Table the $k \times k$ square with $i,j$ entry $i \diamond j$ (assuming a known order). This single operation defines a quasigroup if the table is a Latin Square, each symbol appears once in each row and column. Equivalently, for every $a,b$ there are unique $x,y$ with $x \diamond a=b$ and $a \diamond y=b.$ This defines two operations $y=a \backslash b$ and $x=a / b$ with $$b=a \diamond (a \backslash b)=a \backslash (a \diamond b)=(b /a) \diamond a=(b \diamond a) / a.$$

If $k$ is a prime power (and perhaps only then) it is possible to have $k-1$ Mutually Orthogonal Latin Squares in that any two give all $k^2$ possible ordered pairs when combined position by position. This can go up to $k+1$ if combined with projection onto the first and second coordinates. That should be possible to describe as $k-1$ or $k+1$ related operations.